Questions: Considere um retângulo ABCD e o quadrilátero PQRS, cujos vértices são os pontos médios de AB, BC, CD e AD. Se o perímetro do quadrilátero PQRS é igual a 20 cm, então a medida da diagonal AC do retângulo ABCD é igual a A) 4 cm. B) 5 cm. C) 8 cm. D) 10 cm.

Solution Steps

We are given a rectangle \(ABCD\) and a quadrilateral \(PQRS\) whose vertices are the midpoints of \(AB\), \(BC\), \(CD\), and \(DA\). The perimeter of \(PQRS\) is 20 cm. We need to find the length of the diagonal \(AC\) of the rectangle \(ABCD\).

Since \(P\), \(Q\), \(R\), and \(S\) are midpoints, \(PQRS\) is a rectangle. The sides of \(PQRS\) are parallel to the diagonals of \(ABCD\) and half their length.

Let the length of \(AB\) be \(a\) and the length of \(BC\) be \(b\). The sides of \(PQRS\) are \(\frac{a}{2}\) and \(\frac{b}{2}\).

The perimeter of \(PQRS\) is given by: \[ 2 \left( \frac{a}{2} + \frac{b}{2} \right) = 20 \] \[ \frac{a}{2} + \frac{b}{2} = 10 \] \[ a + b = 20 \]

The diagonal \(AC\) of rectangle \(ABCD\) can be found using the Pythagorean theorem: \[ AC = \sqrt{a^2 + b^2} \]

Since \(a + b = 20\), we can express \(b\) as \(b = 20 - a\).

\[ AC = \sqrt{a^2 + (20 - a)^2} \] \[ AC = \sqrt{a^2 + 400 - 40a + a^2} \] \[ AC = \sqrt{2a^2 - 40a + 400} \]

To find the minimum value of \(AC\), we can complete the square: \[ AC = \sqrt{2(a^2 - 20a + 200)} \] \[ AC = \sqrt{2((a - 10)^2 + 100)} \] The minimum value occurs when \((a - 10)^2 = 0\), i.e., \(a = 10\).

\[ AC = \sqrt{2(100)} \] \[ AC = \sqrt{200} \] \[ AC = 10\sqrt{2} \]

Final Answer